Is this a gauge symmetry?

1. Imagine a hypothetical action: $$S=\int \left(\frac{\partial}{\partial t}\phi(x,t)\right)^2 d^3x dt$$ Now we have a symmetry of the action: $$\phi(x,t)\rightarrow \phi(x,t)+\chi(x).$$ At time $$t$$, $$\phi(x,t)$$ is a function of $$x$$. But by the transform it can be transformed into any other function of $$x$$. Does this mean that there is only one state at time $$t$$ since all states can be related by a transformation?

2. Compared it with the action: $$S=\frac{m}{2}\int \dot{x}(t)^2 dt.$$ It is invariant under a translation $$x\rightarrow x+a$$. But we wouldn't say all positions of a particle are the same. (Or would we?)

So I am confused as a gauge transform is not supposed to allow you to transform inequivalent states into each other. (How would you fix the gauge in the first example?)

What I'm mainly confused about is that while a translation symmetry is thought of as a "real" symmetry, gauge symmetry is thought of as a fake symmetry. But I can't see the difference. In the second example I presume one would set an arbitary coordinate system by setting $$x(0)=a$$ for example. In the first example one would fix $$\phi(x,0)$$ to be some function.

• For a gauge symmetry, the parameters must be completely arbitrary. It is that arbitrariness makes it a "fake symmetry" in your language, or redundancy. As you have noticed, in your first action, you can only fix the state at a particular time, but not all times. To make sure what symmetries there are and if there are redundancies (gauge symmetries), the best way is to perform a Hamiltonian analysis. – chichi Oct 11 '19 at 2:12
• So, the gauge field must depend on time too? – zooby Oct 11 '19 at 2:23
• The gauge parameters are arbitrary functions of spacetime. – chichi Oct 11 '19 at 4:35

2 Answers

You may well be stumbling on unfortunate notation. Your first action, $$S_1=\int \left(\frac{\partial}{\partial t}\phi(x,t)\right)^2 dx^3 dt$$ symmetric under $$\phi(x,t)\rightarrow \phi(x,t)+\chi(x)$$, is but a decoupled sum of your second one, with the unfortunate variable conflation excised, $$S_2= \int \dot{\phi}(t)^2 dt$$ invariant under a translation $$x\rightarrow x+\chi$$, a global symmetry.

That is, $$S_1=\int d^3x ~ S_2(x)$$, where x is but a bland "flavor" label multiplexing the arguments of each copy of $$S_2$$. At no point do you get a gauge symmetry, that is a transformation parameter dependent on a variable with respect to which derivatives are taken in the action to link up the different pieces of the integral.

Your first action has an infinity of global symmetries. Your insinuating a gauge symmetry by virtue of the mere "name" of a label is unwarranted.

• Ah, I see. So it is an infinite set of global symmetries. – zooby Oct 11 '19 at 1:59
1. OP's first Lagrangian density is $${\cal L}~=~\frac{1}{2}(\partial_t \phi)^2 \tag{1a}$$ with a propagating field $$\phi$$, i.e. with the presence of time derivatives. The complete solution to the EL eq. reads $$\phi(x,t)~\approx~f_1(x)t+f_2(x).\tag{1b}$$ Given an initial time surface $$\{t=t_i\}$$, the functions $$f_1(x)$$ and $$f_2(x)$$ are Cauchy data that uniquely specify a physical solution (1b). Consider the infinitesimal transformation $$\delta\phi(x,t)~=~\varepsilon_1(x)t+\varepsilon_2(x).\tag{1c}$$ The infinitesimal transformation (1c) is a quasi-symmetry of the Lagrangian density (1a) $$\delta{\cal L}~=~\frac{d[\phi \varepsilon_1]}{dt}.\tag{1d}$$ It is not a gauge-symmetry since it maps between physically distinguishable configurations.

2. For comparison, consider the Lagrangian density$$^{\dagger}$$ $${\cal L}~=~\frac{1}{2}(\partial_x \phi)^2 \tag{2a}$$ with an auxiliary unphysical field $$\phi$$, i.e. with the absence of time derivatives. The complete solution to the EL eq. reads $$\phi(x,t)~\approx~f_1(t)x+f_2(t).\tag{2b}$$ Consider the infinitesimal transformation $$\delta\phi(x,t)~=~\varepsilon_1(t)x+\varepsilon_2(t).\tag{2c}$$ The infinitesimal transformation (2c) is a quasi-symmetry of the Lagrangian density (2a) $$\delta{\cal L}~=~\frac{d[\phi \varepsilon_1]}{dx}.\tag{2d}$$ It is a gauge-symmetry.

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$$^{\dagger}$$ Let us for simplicity consider a 1+1D spacetime.